# Solve the initial value problem displaystyle{leftlbracebegin{matrix}{y}text{}+{16}{y}= cos{{left({4}{t}right)}}{y}{left({0}right)}={1}{y}'{left({0}right)}={1}end{matrix}right.} using the Laplace transform.

Question
Laplace transform
Solve the initial value problem $$\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}$$ using the Laplace transform.

2020-10-24
Step 1
Given: $$\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Step 2
Explanation:
$$\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Taking Laplace transform on both sides
$$\displaystyle{L}{\left[{y}{''}+{16}{y}\right]}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}$$
$$\displaystyle{L}{\left({y}\text{}{\left({t}\right)}\right)}+{L}{\left({16}{y}{\left({t}\right)}\right)}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}\ldots{\left({i}\right)}$$
We know that if $$L(y(t))=y(s)$$
then $$\displaystyle{L}{\left({y}{''}{\left({t}\right)}\right)}={s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}$$
$$\displaystyle{L}{\left[ \cos{{\left({a}{t}\right)}}\right]}=\frac{s}{{{s}^{2}+{a}^{2}}}$$
Using these in equation (i), we get
$$\displaystyle{s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{16}{y}{\left({s}\right)}=\frac{s}{{{s}^{2}+{4}^{2}}}$$
Using the given initial conditions $$y(0)=1, y'(0)=1$$ then
$$\displaystyle{\left({s}^{2}+{16}\right)}{y}{\left({s}\right)}-{s}{\left({1}\right)}-{1}=\frac{s}{{{s}^{2}+{4}^{2}}}$$
$$\displaystyle{\left({s}^{2}+{4}^{2}\right)}{y}{\left({s}\right)}={1}+{s}+\frac{s}{{{s}^{2}+{4}^{2}}}$$
$$\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}{\left({s}^{2}+{4}^{2}\right)}}}$$
$$\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}^{2}}}$$
Taking inverse Laplace transform on both sides
$$\displaystyle{L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}$$
$$\displaystyle{y}{\left({t}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}$$
$$\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{{2}\cdot{4}}}{t} \sin{{\left({4}{t}\right)}}$$
$$\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{8}}{t} \sin{{\left({4}{t}\right)}}$$

### Relevant Questions

use the Laplace transform to solve the given initial-value problem.
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$$y"+y=g(t) , y(0)=-4 , y'(0)=0$$
where $$g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}$$
Y(s)-?
With the aid of Laplace Transform, solve the Initial Value Problem
$${y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}$$
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}-{6}{y}'+{13}{y}={0},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}=-{9}$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
Use the Laplace transform to solve the given initial-value problem
$${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Use Laplace transform to solve the following initial-value problem
$$y"+2y'+y=0$$
$$y(0)=1, y'(0)=1$$
a) \displaystyle{e}^{{-{t}}}+{t}{e}^{{-{t}}}\)
b) \displaystyle{e}^{t}+{2}{t}{e}^{t}\)
c) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{t}\)
d) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
e) \displaystyle{2}{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
f) Non of the above
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$